A number theory problem by Sravan Chinta

Number Theory Level 4

$Find\quad the\quad number\quad of\quad solutions\quad of\quad (a,b,c)\quad such\quad that\quad \\ \frac { (b+c) }{ a } \quad =\quad \frac { a }{ b } +\frac { a }{ c } \quad \& \quad a,b,c\quad are\quad non-negative\quad integers\\ \& \quad a,b,c<10$

The answer is 17.

1 solution

Darwin Ajah
Dec 27, 2014

(b+c) * bc = a^2 * (b+c)

bc = a^2

a = {1,2,3,4,5,6,7,8,9}

b and c

a = 1 (1 solution)

1 and 1

a = 2 (3 solution)

1 and 4

2 and 2

4 and 1

a = 3 (3 solution)

1 and 9

3 and 3

9 and 1

a = 4 (3 solution)

2 and 8

4 and 4

8 and 2

a = 5 (1 solution)

5 and 5

a = 6 (3 solution)

4 and 9

6 and 6

9 and 4

a = 7 (1 solution)

7 and 7

a = 8 (1 solution)

8 and 8

a = 9 (1 solution)

9 and 9

so there are 17 solutions for bc = a^2 where a,b,c < 10 and non-negatif integer

i think non-negatif it's mean 0 and positif. if 0 is including then there 18 solution

To your last comment about the 0, none of a, b, or c can be equal to 0 because looking at the original expression, we would get undefined in any case. Thus 17 is correct regardless of the wording of the question.

Michael Yi - 6 years, 5 months ago

A number theory problem by Sravan Chinta

The answer is 17.

1 solution

0 pending reports